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Mathematics · Complex Numbers

JEE Main 2024 — 27 January, Shift 2 — Question 30

Let the complex numbers α and 1α\alpha \,and\,\frac{1}{\alpha } lie o the circles ∣z−z0∣2=4{{\left| z-{{z}_{0}} \right|}^{2}}=4 and ∣z−z0∣2=16{{\left| z-{{z}_{0}} \right|}^{2}}=16 respectively, where z0=1+i{{z}_{0}}=1+i. Then the value of 100∣α∣2{{\left| \alpha \right|}^{2}} is….

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

∣z−z0∣2=4⇒(αz0)(α‾z‾0)=4⇒αα‾αz‾0−z0α‾+∣z0∣2=4⇒∣z0∣2αz‾0z0α‾=2..............(1)∣zz0∣2=16⇒(1α‾z0)(1α−z‾0)=16⇒(1−α‾z0)(1−αz‾0)=16∣α∣2⇒1−α‾z0−αz‾0+∣α∣2∣z0∣2=16∣α∣2⇒1−α‾z0−αz‾0=14∣α∣2.............(2)\begin{aligned}& {{\left| z-{{z}_{0}} \right|}^{2}}=4 \\ & \Rightarrow \left( \alpha {{z}_{0}}\right)\left( \overline{\alpha}{{\overline{z}}_{_{0}}}\right)=4\\&\Rightarrow\alpha\overline{\alpha}\alpha{{\overline{z}}_{_{0}}}-{{z}_{0}}\overline{\alpha }+{{\left| {{z}_{0}} \right|}^{2}}=4 \\ & \Rightarrow {{\left| {{z}_{0}}\right|}^{2}}\alpha{{\overline{z}}_{_{0}}}{{z}_{0}}\overline{\alpha}=2..............(1)\\&{{\left|z{{z}_{0}}\right|}^{2}}=16\\&\Rightarrow\left(\frac{1}{\overline{\alpha}}{{z}_{0}}\right)\left(\frac{1}{\alpha }-{{\overline{z}}_{_{0}}} \right)=16 \\& \Rightarrow \left( 1-\overline{\alpha }{{z}_{0}}\right)\left( 1-\alpha {{\overline{z}}_{_{0}}} \right)=16{{\left| \alpha \right|}^{2}} \\ & \Rightarrow 1-\overline{\alpha }{{z}_{0}}-\alpha {{\overline{z}}_{_{0}}}+{{\left| \alpha \right|}^{2}}{{\left| {{z}_{0}} \right|}^{2}}=16{{\left| \alpha \right|}^{2}} \\& \Rightarrow 1-\overline{\alpha }{{z}_{0}}-\alpha {{\overline{z}}_{_{0}}}=14{{\left| \alpha \right|}^{2}}.............(2) \\\end{aligned}

From (1) and (2)

⇒5∣α∣2=1\Rightarrow 5{{\left| \alpha \right|}^{2}}=1

 ⇒100∣α∣2=20\ \Rightarrow 100{{\left| \alpha \right|}^{2}}=20 \\

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Mathematics
Chapter
Complex Numbers
Topic
Conjugate of complex numbers & properties